Mathematical Tool of Quantum Mechanics | Quantum Science Philippines
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Mathematical Tool of Quantum Mechanics

 


Problem:  

Consider the set of all entities of the form (a, b, c) where the entries are real numbers. Addition and scalar multiplication are defined as follows: 

\begin{array}{ccc} (a, b, c) + (d, e, f) &=& (a+d, b+e, c+f)\\ a(a, b, c) &=& (aa, ab,ac) \end{array}

i.) Write down the null vector and inverse of (a, b, c).

ii.) Show that vectors of the form (a, b, 1) do not form a vector space.



 

Solution: 

i.) Let (x, y, z) be the null vector of (a, b, c), such that

(a, b, c) + (x, y, z) = (a, b, c)

Then as defined,

(a+x, b+y, c+z) = (a, b, c)

and by comparison,

\begin{array}{ccc} a+x &=& a \\ b+y &=& b \\ c+z &=& c \end{array}

If we are to solve these equations for the values of x, y and z, we could get

\begin{array}{ccc} x &=& 0 \\ y &=& 0 \\ z &=& 0 \end{array}.

Therefore, the null vector is given as

\mathbf{(x, y, z)} = \mathbf{(0, 0, 0)}.

 

Now, we let the inverse of  (a, b, c) be  (\alpha, \beta, \gamma), such that following the definition of an inverse vector we have

(a, b, c) + ( \alpha , \beta , \gamma) = (0, 0, 0)

( a + \alpha , b + \beta , c+ \gamma ) = (0, 0, 0)

Following the same process as before, if we compare we can obtain

\begin{array}{ccc} a+\alpha &=& 0 \\ b+\beta &=& 0 \\ c+\gamma &=& 0 \end{array}

Solving for \alpha, \beta and \gamma,

\begin{array}{ccc} \alpha &=& -a \\ \beta &=& -b \\ \gamma &=& -c \end{array}.

Therefore, the inverse vector of (a, b, c) is

\mathbf{( \alpha, \beta, \gamma )} = \mathbf{( -a, -b, -c)}.

 

 

ii.) The vector  (a, b, 1) do not form a vector space because:

a.) it violates the closure under addition,

\begin{array}{ccc} ( a, b, 1) + ( c, d, 1) &=& ( a+c, b+d, 1+1) \\ &=& \mathbf{( a+c, b+d, 2) \not\in (a,b, 1)} \end{array}.

               b.) it violates the closure under scalar multiplication,

\lambda (a, b, 1) = \mathbf{( \lambda a, \lambda b, \lambda ) \not\in ( a, b, 1)}.

               c.) there is NO null vector

(0, 0, 0) \not\in (a, b, 1)

               d. an additive inverse does not exist

( -a, -b, -1) \not\in (a, b, 1).

 

 

By: Joshua J. Ordeniza, MS Physics Student, MSU-IIT

 


 

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